Let the line $L_1$ be parallel to the vector $-3\hat{i} + 2\hat{j} + 4\hat{k}$ and pass through the point $(2, 6, 7)$, and the line $L_2$ be parallel to the vector $2\hat{i} + \hat{j} + 3\hat{k}$ and pass through the point $(4, 3, 5)$. If the line $L_3$ is parallel to the vector $-3\hat{i} + 5\hat{j} + 16\hat{k}$ and intersects the lines $L_1$ and $L_2$ at the points $C$ and $D$, respectively, then $\left|\overrightarrow{CD}\right|^2$ is equal to:

Question

Accepted Answer

A. 1. Write $L_1$ and $L_2$ in parametric form using parameters $t$ and $s$ respectively.

$$
L_1: \frac{x-2}{-3}=\frac{y-6}{2}=\frac{z-7}{4}=t,\qquad L_2: \frac{x-4}{2}=\frac{y-3}{1}=\frac{z-5}{3}=s
$$

2. So the points $C$ (on $L_1$) and $D$ (on $L_2$) can be expressed as

$$
C=(-3 t+2,2 t+6,4 t+7),\qquad D=(2 s+4, s+3,3 s+5)
$$

3. The direction ratios of $\overrightarrow{CD}$ are found by subtracting coordinates.

$$
\overrightarrow{CD}=(2 s+3 t+2,\ s-2 t-3,\ 3 s-4 t-2)
$$

4. Since $L_3$ (and hence $CD$) is parallel to $-3\hat{i}+5\hat{j}+16\hat{k}$, the components of $\overrightarrow{CD}$ must be proportional to $(-3,5,16)$. Equating ratios gives two independent linear equations in $s$ and $t$.

$$
13 s+9 t+1=0,\qquad 41 s+36 t+26=0
$$

5. Solving this system of two equations simultaneously,

$$
s=2,\qquad t=-3
$$

6. Substituting back gives the actual coordinates of $C$ and $D$.

$$
C=(11,0,-5),\qquad D=(8,5,11)
$$

7. Using the distance formula to compute $CD^2$,

$$
CD^2=(11-8)^2+(0-5)^2+(-5-11)^2=9+25+256=290
$$

Hence, the answer is **Option A**, $290$.